Solubility Product Constant (Ksp) and the Common-Ion Effect for Calcium Iodate, a Salt of Limited Solubility

Purpose

Determine the solubility product constant (Ksp) for a sparingly soluble salt. Study the effect on the aqueous solubility equilibrium of this salt in the presence of an additional amount of one of the ions that it contains (in this case, the effect of added Ca2+).

Introduction

This experiment deals with saturated aqueous solutions of salts with limited solubility in water. Under ordinary conditions, it might be fair to call the material you'll use today as insoluble, or at best slightly soluble. At such conditions of limited aqueous solubility, the solid salt and its ions in solution are in a state of dynamic equilibrium between the solid phase and the solution phase. Concentrations of ions in solution are exceedingly low, and care must be taken to obtain good results, when performing the experimental procedure outlined shortly. As an example of a slightly soluble salt, consider a saturated solution of silver sulfate, Ag2SO4. In this example, the equilibrium lies far to the left, owing to the fact that silver sulfate is only slightly soluble in water: - + 2  Ag2SO4(s) <-===== > 2Ag (aq) + SO4 (aq)

The corresponding equilibrium constant expression for this system is:

+ 2 2− [Ag ] [SO4 ] K = ------[Ag2SO4]

Since the solid component has a near-constant concentration, it is not used in the equilibrium expression, simplifying it to the solubility product:

+ 2 2− K = [Ag ] [SO4 ]

The new symbol Ksp is called the solubility product constant. Most solubility product constants have very −5 small values, as is the case for silver sulfate, where Ksp = 1.1 x 10 .

−5 + 2 2− 1.1 x 10 = [Ag ] [SO4 ]

In 1 liter of a saturated, aqueous solution of silver sulfate, the molar concentration of silver ion can be found by letting x equal the sulfate ion concentration, 2x the silver ion concentration, and substituting into the solubility product equation the known value for Ksp:

− 1.1 x 10−5 = [Ag+]2 [SO 2 ] 4 1.1 x 10−5 = [2x]2 [x]

1.1 x 10−5 = 4 x3

2.75 x 10−6 = x3

2− −2 x = [SO4 ] = 1.4 x 10 M

+ 2x = [Ag ] = 2.8 x 10−2 M Background

The equilibrium process in this experiment is a saturated aqueous solution of calcium iodate, Ca(IO3)2. The relevant solubility equation and solubility product expression, are both shown below.

- 2+ − Ca(IO3)2(s) <-===== > Ca (aq) + 2IO3 (aq)

2+ − 2 Ksp = [Ca ] [IO3 ]

For a saturated solution of calcium iodate, if you can determine either the molar concentration of calcium ion, or the molar concentration iodate ion, the solubility product constant can be found using the reverse of the process shown above. There was found the silver ion concentration, in a saturated aqueous solution, from a known value for Ksp. In other words, if the calcium ion concentration in today's experiment was found to be 0.1 M, you could immediately say the concentration of iodate ion must be half that value, or 0.05 M, according to the stoichiometry of the solubility equation given above. The solubility product constant could then be found with simple arithmetic. In this experiment, the iodate ion concentration of a saturated calcium iodate solution will be found via a redox titration with sodium thiosulfate, Na2S2O3. Also, the effect of adding a source of additional iodate ion on the solubility of calcium iodate will be determined. This is accomplished via redox titration of a second calcium iodate solution containing an additional source of aqueous iodate ion. It is possible to predict the result qualitatively upon the calcium iodate equilibrium, before actually performing the experiment. According to LeChâtelier's principle, adding an additional source of iodate ion (besides that of calcium iodate itself) should cause the solubility equilibrium to shift to the left, favoring more solid calcium iodate. This is an example of the common-ion effect. The common-ion effect, in this experiment, should lead to a reduced solubility of calcium iodate, and a corresponding change in the solubility product constant. Details of the Redox Titration of Iodate Ion

Two solutions of calcium iodate will be used. One is in pure deionized water. The other is in a water solution containing 0.0100M KIO3. You will measure the concentration of the iodate ion in each solution. - The iodate ion concentration will be determined using an indirect redox titration. The IO3 ion is an - oxidizing agent. It reacts with iodide ions, I , to produce iodine molecules, I2. The iodine produced is then

titrated with a standardized sodium thiosulfate solution, Na2S2O3. Iodine in water, particularly when there - are I ions present, gives a deep reddish brown to a pale yellow color to the solution, depending on its concentration. If a starch solution is added to the iodine solution when the concentration of iodine is low, and the solution appears yellow, the iodine and the starch form an intensely blue colored complex. The disappearance of the blue color as sodium thiosulfate is added gives the endpoint in the titration. The relevant reaction equations are summarized as follows.

− − + IO3 (aq) + 5I (aq) + 6H3O (aq) ------> 3I2(aq) + 9H2O(l) RX #1

This step, which occurs after adding both solid KI, and aqueous acid, to aliquots of saturated iodate solutions, has the net effect of converting iodate ions to aqueous iodine. Thiosulfate ion then reacts with aqueous iodine according to: 2− − 2− I2(aq) + 2S2O3 (aq) ------> 2I (aq) + S4O6 (aq) RX #2

The net titration reaction can be obtained by combining the two reactions above, then balancing for mass and charge: − − + IO3 (aq) + 5I (aq) + 6H3O (aq) ------> 3I2(aq) + 9H2O(l)

2− − 2− 3I2(aq) + 6S2O3 (aq) ------> 6I (aq) + 3S4O6 (aq) ------− 2− + − 2− IO3 (aq) + 6S2O3 (aq) + 6H3O (aq) ------> I (aq) + 3S4O6 (aq) + 9H2O(l)

In reaction (1), the I- is provided by adding solid KI in excess, and the H+ is provided by adding HCl - solution. In reaction (1), the I2 produced comes from the IO3 present. There are 3 moles of I2 produced for - 2- every 1 mole of IO3 present. In reaction (2), 2 moles of S2O3 are required for every mole of I2 present. The 2- - net result is that for every 6 moles of S2O3 used in the titration, there must have been 1 mole of IO3 originally present. You will use the molarity of the sodium thiosulfate solution to calculate the molarity of the iodate in the calcium iodate solution.

It is important to notice the molar ratio of iodate ion to thiosulfate ion in the above complete, balanced titration reaction. Ask your instructor if the thiosulfate solution available for you to use has been standardized, or whether you will have to standardize the thiosulfate solution as part of the experimental procedure. Aliquots from prepared, saturated solutions of calcium iodate in water, with and without additional calcium ion, will be titrated according to the reaction chemistry of the equations outlined above.

Experimental Notes • It is important to handle the saturated iodate solutions with great care, so as not to allow any of the solid calcium iodate into you titrations. This will obviously lead to a higher value for concentration of iodate than is actually present in the saturated solutions. • It is recommended that you perform 2 titrations of unknown calcium iodate after standardizing the thiosulfate solution. The first of these should be a 'scout', where 1 mL portions of thiosulfate solution are added quickly to determine the approximate volume needed to reach equivalence. • Titrated solutions, and any left over thiosulfate or iodate solutions, are safe to dilute and rinse down the sinks.

Experimental Procedure

Note: it is important for the Erlenmeyer flasks and other glassware you'll use to be as clean as possible when performing these redox titrations. Please take a moment to at least give them a thorough rinsing--they need not be completely dry before proceeding.

Saturated solutions of calcium iodate in water, and calcium iodate with added iodate ion, are available in the hoods. Please take some care to not disturb the solid calcium iodate present in the bottles when obtained your portions of solutions to use.

Obtain approximately 40 mL of the two different calcium iodate solutions in small beakers. One is labeled "Calcium Iodate, Aqueous" and the other "Calcium Iodate, Added Calcium Ion". You will eventually use about 150 mL of this standardized thiosulfate solution.

Part 1 - Titration of iodate in saturated calcium iodate solution

Set up a 50 mL buret and stand, fill the buret with standardized sodium thiosulfate solution, and take a reading of initial volume to two decimal places. Be sure and record the exact concentration of the thiosulfate solution that is listed on the reagent bottle. This solution has been standardized by the Preparation Lab, and its concentration should be printed on the label. Pipet 10.0 mL of a calcium iodate saturated solution into a clean 125 mL Erlenmeyer flask. Using a graduated cylinder, add approximately 20 mL of distilled water to this saturated iodate solution, swirl to mix. Dissolve ~ 0.5g of solid KI into the iodate/water solution, then add 10 mL of 1 M HCl. Swirl to mix the contents, obtaining dark red homogeneous solutions. Titrate the resulting brown solution with sodium thiosulfate until the brown color

(I2) is mostly gone and the solution turned pale yellow (not golden). At this point, add 5 mL off a 0.1% starch solution. The titration solutions should be a dark blue-black color. Titrate with standardized thiosulfate solution until a colorless endpoint.

Part 2 - Titration of iodate in saturated calcium iodate solution prepared in 0.0100 M KIO3

Repeat the procedure from Part 1 above using 10 mL of saturated Ca(IO3)2 in 0.0100 M KIO3.

Data Collection

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Part I – Calculation Worksheet

ICE Table

2+ - ↔ Ca (aq) 2 IO3 (aq) Ca(IO3)2

I

C

E

Ksp =

Part II - Calculation Worksheet

ICE Table

2+ - ↔ Ca (aq) 2 IO3 (aq) Ca(IO3)2

I

C

E

Ksp =

Data Summary Table -- Solubility Product Constant/Common-Ion Effect

Part A: Calcium Iodate, No Added Iodate Ion Trial #1 Trial #2

Concentration of Sodium Thiosulfate Solution (M) ______

Volume of Calcium Iodate Solution Added (L) ______

Final Volume, Thiosulfate Solution (mL) ______

Initial Volume, Thiosulfate Solution (mL) ______

Total Volume, Thiosulfate Solution (mL) ______

Total Volume, Thiosulfate Solution (L) ______

Moles Sodium Thiosulfate Used ______

Moles Iodate ______

Equilibrium Concentration of Iodate Ion (M) ______

Equilibrium Concentration of Calcium Ion (M) ______

Molar Solubility of Calcium Iodate (M) ______

Ksp, Calcium Iodate ______

Average For Ksp, Calcium Iodate ______Data Tables -- Solubility Product Constant/Common-Ion Effect

Part B: Calcium Iodate, Added Calcium Ion Trial #1 Trial #2

Concentration of Sodium Thiosulfate Solution (M) ______

Volume of Calcium Iodate Solution Added (L) ______

Final Volume, Thiosulfate Solution (mL) ______

Initial Volume, Thiosulfate Solution (mL) ______

Total Volume, Thiosulfate Solution (mL) ______

Total Volume, Thiosulfate Solution (L) ______

Moles Sodium Thiosulfate ______

Moles Iodate ______

Equilibrium Concentration of Iodate Ion (M) ______

Equilibrium Concentration of Calcium Ion (M) ______

Molar Solubility of Calcium Iodate ______(added Ca2+ ion source) Average Molar Solubility of Calcium Iodate ______(added Ca2+ ion source)

Questions

1) Compare [Ca2+] from Parts 1 and 2 (note: this is the solubility). Is Le Châtelier’s law followed? Explain.

2) Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO3)2. Will a precipitate of PbI2 form? Assume 1 drop = 0.05 mL

2+ - -9 PbI2(s) ↔ Pb (aq) + 2 I (aq) Ksp = 7.1 x 10